# Question 2cc18

Feb 27, 2017

$2 {\text{HCl"_ ((aq)) + "Ca"("OH")_ (2(s)) -> "CaCl"_ (2(aq)) + 2"H"_ 2"O}}_{\left(l\right)}$

#### Explanation:

The idea here is that all the atoms that are present on the reactants' side must also be present on the products' side.

So, you know that the unbalanced chemical equation looks like this

${\text{HCl"_ ((aq)) + "Ca"("OH")_ (2(s)) -> "CaCl"_ (2(aq)) + "H"_ 2"O}}_{\left(l\right)}$

It's always a good idea to start with an inventory of what you have in the unbalanced chemical equation

color(white)(aacolor(blue)(ul(color(black)("the reactants' side")))aaaaaaaaaaaacolor(blue)(ul(color(black)("the products' side")))aaaa

color(white)(aaaaaaaaacolor(black)(3 xx "H")aaaaaaaaaaaaaaaaaaaaaaacolor(black)(2 xx "H")aaaaacolor(red)(xx)

color(white)(aaaaaaaaacolor(black)(1 xx "Ca")aaaaaaaaaaaaaaaaaaaaaacolor(black)(1 xx "Ca")aaaaacolor(darkgreen)(sqrt())

color(white)(aaaaaaaaacolor(black)(1 xx "Cl")aa.aaaaaaaaaaaaaaaaaaaacolor(black)(2 xx "Cl")aaaaacolor(red)(xx)

color(white)(aaaaaaaaacolor(black)(2 xx "O")aaaaaaaaaaaaaaaaaaaaaaacolor(black)(1 xx "O")aaaaacolor(red)(xx)

Start by balancing the chlorine atoms. You have $1$ chlorine atom on the reactants' side and $2$ on the products' side, so add a coefficient of $2$ to hydrochloric acid

$2 {\text{HCl"_ ((aq)) + "Ca"("OH")_ (2(s)) -> "CaCl"_ (2(aq)) + "H"_ 2"O}}_{\left(l\right)}$

You now have

color(white)(aacolor(blue)(ul(color(black)("the reactants' side")))aaaaaaaaaaaacolor(blue)(ul(color(black)("the products' side")))aaaa

color(white)(aaaaaaaaacolor(black)(4 xx "H")aaaaaaaaaaaaaaaaaaaaaaacolor(black)(2 xx "H")aaaaacolor(red)(xx)

color(white)(aaaaaaaaacolor(black)(1 xx "Ca")aaaaaaaaaaaaaaaaaaaaaacolor(black)(1 xx "Ca")aaaaacolor(darkgreen)(sqrt())

color(white)(aaaaaaaaacolor(black)(2 xx "Cl")aa.aaaaaaaaaaaaaaaaaaaacolor(black)(2 xx "Cl")aaaaacolor(darkgreen)(sqrt())

color(white)(aaaaaaaaacolor(black)(2 xx "O")aaaaaaaaaaaaaaaaaaaaaaacolor(black)(1 xx "O")aaaaacolor(red)(xx)

Now, notice that you can balance the hydrogen atoms and the oxygen atoms by adding a coefficient of $2$ to the water

$2 {\text{HCl"_ ((aq)) + "Ca"("OH")_ (2(s)) -> "CaCl"_ (2(aq)) + 2"H"_ 2"O}}_{\left(l\right)}$

You now have

color(white)(aacolor(blue)(ul(color(black)("the reactants' side")))aaaaaaaaaaaacolor(blue)(ul(color(black)("the products' side")))aaaa

color(white)(aaaaaaaaacolor(black)(4 xx "H")aaaaaaaaaaaaaaaaaaaaaaacolor(black)(4 xx "H")aaaaacolor(darkgreen)(sqrt())

color(white)(aaaaaaaaacolor(black)(1 xx "Ca")aaaaaaaaaaaaaaaaaaaaaacolor(black)(1 xx "Ca")aaaaacolor(darkgreen)(sqrt())

color(white)(aaaaaaaaacolor(black)(2 xx "Cl")aa.aaaaaaaaaaaaaaaaaaaacolor(black)(2 xx "Cl")aaaaacolor(darkgreen)(sqrt())#

$\textcolor{w h i t e}{a a a a a a a a a \textcolor{b l a c k}{2 \times \text{O")aaaaaaaaaaaaaaaaaaaaaaacolor(black)(2 xx "O}} a a a a a \textcolor{\mathrm{da} r k g r e e n}{\sqrt{}}}$

And there you have it, the chemical equation is now balanced.